Question

`5` different balls are placed in `5` different boxes randomly. Find the probability that exactly two boxes remain empty. Given each box can hold any number of balls.
A. `(24)/(125)`
B. `(12)/(25)`
C. `(96)/(125)`
D. None of these

Answer 1

Correct Answer - B
`(b)` Each ball can be placed in 5 ways.
`:.` Total number of ways, `n(S)=5^(5)`
`2` empty boxes, can be selected in `"^(5)C_(2)` ways and `5` balls can be placed in the remaining `3` boxes in groups of `221` or `311` in
`3![(5!)/(2!2!2!)+(5!)/(3!2!)]=150` ways
`:.` Favorable cases `n(A)="^(5)C_(2)*150`
Hence `P(A)="^(5)C_(2)*(15)/(5^(5))=(60)/(125)=(12)/(25)`

Answer 2

Answer: (B) \(\frac{12}{25}\)

We have, the number elements in sample space = n(S) =5⁵ as placing 5 different marbles in 5 different boxes.

For computing favourable outcomes, 2 boxes which are to remain empty, can be selected in ⁵C₂ ways and 5 marbles can be placed in the remaining 3 boxes in groups of 221 or 311 in

\(3![\frac{5!}{2!2!2!}+\frac{5!}{3!2!}]\)

\(=(3\times 2\times 2)[\frac{5\times 4\times 3\times 2!}{2\times 2\times 2 \times 2}+\frac{5\times 4 \times 3!}{3!\times2}]\)

= 6 × [ 15 + 10 ] 

= 150 ways 

Thus, number of elements in favourable event, n( A ) = ⁵C₂ × 150

Hence, P(E) = \(\frac{n(A)}{n(S)}=^5C_2\times \frac{150}{5^5}=\frac{60}{125}=\frac{12}{25}\)